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ki77a [65]
3 years ago
7

Which point is on the line that passes through (0,6) and is parallel to the given line? plzz helppp!

Mathematics
1 answer:
Grace [21]3 years ago
5 0

Points on given line = (-12,-2) and (0,-4) because you can see them on the graph.  Then draw a parallel line thru (0,6)

To get from (0,-4) to (0,6) your x stays constant and your y coordinate increased by 10.  So your new point will do the same in relation to (-12,-2)  the x will stay constant at -12 and your y will increase by 10 to +8.  

So the answer is A (-12,8)

You can check this because parallel lines have the same slope so

y2-y1/x2-x1 should be equal for both lines.

Line 1:  -4 - (-2) / 0 - (-12) = -2/12 = -1/6

Line 2:  6 - 8 / 0 - (-12) = -2/12 = -1/6


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kirill115 [55]

No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.

Consider polynomial equations x^2 = 3 and x^2 = 1.

Obviously they have real solutions.

Subtract the two polynomial equations:

(x^2 - x^2) = (3 - 1)

0 = 2...

We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.

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3 0
2 years ago
m∠AOB = 6x + 5, m∠BOC = 4x - 2, m∠AOC = 8x + 21
AnnZ [28]
The sum of angle AOB and angle BOC is equal tot he angle of AOC so

6x+5 +4x-2 = 8x+21

10x+3=8x+21

Subtract three and 8x from both sides

2x=18

Divide by two

x=9

Hope this helped!
6 0
3 years ago
Please show the work, i’ll mark as brainliest
vazorg [7]

Answer:

hope it helps you

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6 0
2 years ago
Write the equation of the line that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5)
Sedbober [7]

Answer:   \bold{y=-\dfrac{1}{3}x-1}

<u>Step-by-step explanation:</u>

(4, 1) & (2, -5)

First, find the slope (m) and then the perpendicular (opposite reciprocal) slope:

m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\\m=\dfrac{-5-1}{2-4} = \dfrac{-6}{-2}=3\quad \rightarrow \quad m_{\perp}=-\dfrac{1}{3}\\

Next, find the midpoint of (4, 1) and (2, -5):

Midpoint=\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{4+2}{2},\dfrac{1-5}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{6}{2},\dfrac{-4}{2}\bigg)\\\\\\.\qquad \qquad=(3, -2)

Lastly, input the perpendicular slope and the midpoint into the Point-Slope formula to find the equation of the line:

y - y_1 = m_{\perp}(x - x_1)\\\\y - (-2) = -\dfrac{1}{3}(x - 3)\\\\y + 2=-\dfrac{1}{3}x +1\\\\y =-\dfrac{1}{3}x - 1\\


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2 years ago
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goldenfox [79]

Answer:eirther C or D not sure

Step-by-step explanation:

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